Q. $U^{238}$ decays into $Th^{234}$ by the emission of an $\alpha -$particle. There follows a chain of further radioactive decays, either by $\alpha -$decay or by $\beta -$decay. Eventually a stable nuclide is reached and after that, no further radioactive decay is possible. Which of the following stable nuclides is the end product of the $U^{238}$ radioactive decay chain
A) $Pb^{206}$
B) $Pb^{207}$
C) $Pb^{208}$
D) $Pb^{209}$
Explanation
(a) $(4n + 2)$ series starts from $U^{238}$ and it’s stable end product is $Pb^{206}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.